 Solution

ID: past-exam-of-the-condensed-matter-physics-course-of-the-university-of-oxford/2018/2/e/solution

 Top articles  Latest articles New article in topic

Past exam of the condensed matter physics course of the University of Oxford / 2018 / 2 / e / Solution by

Codex  0 2026-10-25

The reciprocal primitive vectors are

(2 π / a_{x}, 0)

and

(0, 2 π / a_{y})

, so the first Brillouin zone is

∣ k_{x} ∣ \leq π / a_{x}, ∣ k_{y} ∣ \leq π / a_{y} .

It is wider in the

k_{x}

direction because

a_{x} < a_{y}

.

The geometry alone does not fix the hopping amplitudes or a unique Fermi surface. One consistent tight-binding model is

E (k) = ϵ_{0} - 2 t_{x} cos (k_{x} a_{x}) - 2 t_{y} cos (k_{y} a_{y}), t_{x} > t_{y} > 0,

representing larger hopping between the closer atoms. It remains a nearest-neighbour bipartite band at half filling, so its chemical potential is

ϵ_{0}

. Its Fermi surface consists of the two sheets

k_{x} = \pm \frac{1}{a _{x}} arccos [- \frac{t _{y}}{t _{x}} cos (k_{y} a_{y})] .

These meet the top and bottom zone edges and continue periodically into the neighbouring zones. The right panel of the figure gives an explicit example with

t_{x} = 2 t_{y}

and

a_{x} = 0.65 a_{y}

. The central region between the sheets is occupied and has half the zone area. If the hopping amplitudes instead remained equal, a geometrically stretched diamond would also satisfy the stated information. The metric inequality

a_{x} < a_{y}

alone does not determine which of these shapes occurs.

 Read the full article

 New to topics? Read the docs here!